1. 16.360 Lecture 6 Last lecture: Wave propagation on a Transmission line Characteristic impedance Standing wave and traveling wave Lossless transmission line Reflection coefficient

2. 16.360 Lecture 6 Wave equations d² i(z)/ dz² -  ² i(z) = 0, (13) d² V(z)/ dz² -  ² V(z) = 0, (10)  =  + j ,  = Re ( R’ + j  L’) (G’+ j  C’),  = Im ( R’ + j  L’) (G’+ j  C’), V(z) = V 0 (14) + e -z-z + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz   0,  0, + Z 0  I0I0 + V0V0 = ( R’ + j  L’) (G’+j  C’)

3. 16.360 Lecture 6 lossless transmission line :  =  + j ,  = 0,  =  L’C’ Z0Z0 = L’ C’ = 2  /  =  L’C’ 1 Vp =  /  = L’C’ 1 L’C’ =  = rrrr c Vp = L’C’ 1  =  =  

4. 16.360 Lecture 6 For TEM transmission line : Vp = L’C’ 1 L’C’ =  =  1  =  L’C’ =   = rrrr c Z0Z0 = L’ C’ summary : V(z) = V 0 + + - V0V0 e jzjz i(z) = I 0 + e -j  z + - I0I0 e jzjz e = rrrr c Vp = L’C’ 1  =  =  

5. 16.360 Lecture 6 Voltage reflection coefficient : Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi i(z) = V(z) = V 0 + + - V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 - V0V0 + V0V0 = ZLZL Z0Z0 - ZLZL Z0Z0 +    1.|  |  1, how to prove it? 2.If Z L = Z 0,  = 0. Impedance match, no reflection from the load Z L.

6. 16.360 Lecture 6 Today Standing wave Input impedance i(z) = V(z) = V 0 + + - V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 - V0V0 + V0V0 with  = i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |V(z)| = | V 0 | | | + e -j  z |||| e jzjz + e jrjr = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2

7. 16.360 Lecture 6 Standing wave i(z) = V(z) = V 0 + + - V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 - V0V0 + V0V0 with  = i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |V(z)| = | V 0 | | | + e -j  z |||| e jzjz + e jrjr = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2

8. 16.360 Lecture 6 Standing wave i(z) = V(z) = V 0 ( ) - + +  e jzjz e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |i(z)| = | V 0 | /|Z0|| | + e -j  z |||| e jzjz - e jrjr = | V 0 |/|Z 0 | [ 1+ |  |² - 2|  |cos(2  z +  r )] + 1/2 = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)|

9. 16.360 Lecture 6 Special cases = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| 1.Z L = Z 0,  = 0 = | V 0 | + |V(z)| 2. Z L = 0, short circuit,  = -1 |V(z)| |V0| + - -3 /4- /2- /4 = | V 0 | [ 2 + 2cos(2  z +  )] + 1/2 |V(z)| 2|V0| + - -3 /4- /2- /4

10. 16.360 Lecture 6 Special cases = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| 3. Z L = , open circuit,  = 1 = | V 0 | [ 2 + 2cos(2  z )] + 1/2 |V(z)| 2|V0| + - -3 /4- /2- /4

11. 16.360 Lecture 6 Voltage maximum = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| | V 0 | [ 1+ |  |], + |V(z)| max = when 2  z +  r = 2n . –z =  r /4  + n /2 n = 1, 2, 3, …, if  r <0 n = 0, 1, 2, 3, …, if  r >= 0

12. 16.360 Lecture 6 Voltage minimum = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| | V 0 | [ 1 - |  |], + |V(z)| min = when 2  z +  r = (2n+1) . –z =  r /4  + n /2 + /4 Note: voltage minimums occur /4 away from voltage maximum, because of the 2  z, the special frequency doubled.

13. 16.360 Lecture 6 Voltage standing-wave ratio VSWR or SWR 1 - |  | |V(z)| min S  |V(z)| max = 1 + |  | S = 1, when  = 0, S = , when |  | = 1,

14. 16.360 Lecture 6 An example Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi Voltage probe S = 3, Z 0 = 50 ,  l min = 30cm, l min = 12cm, Z L =?  l min = 30cm,  = 0.6m, S = 3,  |  | = 0.5, Solution: -2  l min +  r = - ,   r = -36º,  , and Z L.

15. 16.360 Lecture 6 Input impudence Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi Zg IiIi Z in (z) = V(z) I(z) = + +  e jzjz e -j  z V0V0 ( ) + -  e jzjz e V0V0 ( ) Z0Z0 = + (1  e j2  z ) - (1  e j2  z ) Z0Z0 + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) =

16. 16.360 Lecture 6 An example A 1.05-GHz generator circuit with series impedance Zg = 10-  and voltage source given by Vg(t) = 10 sin(  t +30º) is connected to a load ZL = 100 +j5-  through a 50- , 67-cm long lossless transmission line. The phase velocity is 0.7c. Find V(z,t) and i(z,t) on the line. Solution: Since, Vp = ƒ, = Vp/f = 0.7c/1.05GHz = 0.2m.  = 2  /,  = 10 .  = (Z L -Z 0 )/(Z L +Z 0 ),  = 0.45exp(j26.6º) + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = = 21.9 + j17.4  V 0 [ exp(-j  l)+  exp(j  l) ] + = Z in (-l) + Zg Z in (-l) Vg

17. 16.360 Lecture 6 Next lecture short circuit line open circuit line quarter-wave transformer matched transmission line